Research Article

Block-by-Block Method for Solving Nonlinear Volterra-Fredholm Integral Equation

Table 4


𝑁 𝑥 = 3 0 , 𝑁 𝑡 = 2 0 , case 1 𝑁 𝑥 = 3 0 , 𝑁 𝑡 = 2 0 , case 2

𝑥 𝑡 Error 𝑥 𝑡 Error

0.9667 0.05 2 . 4 4 0 9 𝑒 8 0.9667 0.05 5 . 5 4 0 𝑒 9
0.9667 0.0625 1 . 0 4 3 8 2 𝑒 7 0.9667 0.0625 1 . 6 7 3 6 𝑒 9
0.9667 0.075 1 . 6 5 9 2 8 𝑒 7 0.9667 0.075 4 . 1 7 4 4 𝑒 8
0.9667 0.0875 4 . 4 8 4 8 8 𝑒 7 0.96670.0875 1 . 9 2 8 7 2 4 5 5 4 𝑒 3
0.9667 0.1 6 . 1 3 8 9 3 𝑒 7 0.9667 0.1 1 . 9 2 8 9 8 7 7 0 2 𝑒 3
0.9667 0.1125 1 . 3 0 4 2 0 𝑒 6 0.9667 0.1125 3 . 1 0 0 9 𝑒 7
0.9667 0.125 1 . 6 5 0 4 4 𝑒 6 0.9667 0.125 5 . 2 6 3 3 𝑒 7
0.9667 0.1375 3 . 0 2 6 9 2 𝑒 6 0.9667 0.1375 8 . 3 0 1 7 𝑒 7
0.9667 0.15 3 . 6 5 0 6 5 𝑒 6 0.9667 0.15 1 . 2 9 2 5 1 3 𝑒 6
0.9667 0.1625 6 . 0 6 5 2 8 𝑒 6 0.9667 0.1625 3 . 4 6 9 2 6 𝑒 3
0.9667 0.175 7 . 0 8 1 8 8 𝑒 6 0.9667 0.175 3 . 4 7 3 9 0 4 7 1 𝑒 3
0.9667 0.1875 1 . 0 9 6 0 6 8 𝑒 6 0.9667 0.1875 3 . 8 2 7 9 9 𝑒 6
0.9667 0.2 1 . 2 5 0 3 1 5 𝑒 5 0.9667 0.2 5 . 3 1 2 5 0 𝑒 6
0.9667 0.2125 1 . 8 3 4 6 5 3 𝑒 5 0.967 0.2125 4 . 4 9 1 3 4 7 3 4 𝑒 3
0.9667 0.225 2 . 0 5 6 4 5 5 𝑒 5 0.967 0.225 4 . 5 0 7 7 5 9 8 5 𝑒 3